Question: A secant line intersects the graph of $g(x)=-2x^2+7$ at two points with $x$ -coordinates $2$ and $t$, where $t\neq 2$. What is the slope of the secant line in terms of $t$ ? Your answer must be fully expanded and simplified.
Answer: We are given that the secant line intersects the graph at $x=2$ and $x=t$. Since these points are on the the graph of $g(x)=-2x^2+7$, we know that they must be $(2,-1)$ and $(t,-2t^2+7)$, respectively. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{-2t^2+7-(-1)}{t-2} \\\\ &=\dfrac{-2t^2+8}{t-2} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} \dfrac{-2t^2+8}{t-2}&=\dfrac{-2(t^2-4)}{t-2} \\\\ &=\dfrac{-2(t-2)(t+2)}{t-2} \\\\ &=-2(t+2)\text{, for }t\neq 2 \end{aligned}$ Since we are given that $t\neq 2$, we can conclude that the slope of the secant line is $-2(t+2)$ or $-2t-4$.